gram matrix for elements of an infinite dimensional hilbert space

Question

I know that a gram matrix is defined as $G = V^*V$ where $G = (\langle x_i , x_j \rangle)_{i=1, j=1}^n$ where $x_i, x_j \in H$ and $H$ is some inner product space, say a Hilbert space. I'm confused about how to think of this definition when we talk about infinite dimensional hilbert spaces, say $L^2[a,b]$. If i take a finite subset $(f_i)_{i=1}^n \subset L^2$, then constructing a matrix of inner products between the elements is straight forward, but what does $V^*V$ actually mean now? Does this definition still hold, how can I decompose $G$?

Answer 1

For separable infinite-dimensional Hilbert spaces $(\mathcal H,\langle\,\cdot,\cdot\,\rangle_{\mathcal H})$ (such as for example $L^2)$, this matrix decomposition still works if one considers matrices of infinite size. Use that every such Hilbert space $\mathcal H$ has a countable orthonormal basis $(g_n)_{n\in\mathbb N}$, so it is isometrically isomorphic to the Hilbert space $\ell_2$ of square-summable sequences via the linear mapping $$ \phi:\mathcal H\to\ell_2\qquad x\mapsto\begin{pmatrix}\langle g_1,x\rangle_{\mathcal H}\\\langle g_2,x\rangle_{\mathcal H}\\\vdots\end{pmatrix} $$ (this can be seen via the basis expansion and Parseval's identity). Now $\phi(x)$ is a vector of infinite length, so we can consider the "$\infty\times n$" matrix $V$ generated by $f_1,\ldots,f_n\in\mathcal H$ via $$ V=\begin{pmatrix}\phi(f_1)&\ldots&\phi(f_n)\end{pmatrix}=\begin{pmatrix} \langle g_1,f_1\rangle&\langle g_1,f_2\rangle&\cdots&\langle g_1,f_n\rangle\\\langle g_2,f_1\rangle&\cdots&\cdots&\langle g_2,f_n\rangle\\\vdots&\cdots&\cdots&\vdots\\ \end{pmatrix}. $$ Then one has $$ V^\dagger V=\begin{pmatrix}(\phi(f_1))^\dagger\\\vdots\\(\phi(f_n))^\dagger\end{pmatrix}\begin{pmatrix}\phi(f_1)&\ldots&\phi(f_n)\end{pmatrix}=\big( (\phi(f_i))^\dagger\phi(f_j)\big)_{i,j=1}^n\in\mathbb C^{n\times n} $$ with $$ (\phi(f_i))^\dagger\phi(f_j)=\langle\phi(f_i),\phi(f_j)\rangle_{\ell_2}=\sum_{k\in\mathbb N}\overline{\langle g_k,f_i\rangle}\langle g_k,f_j\rangle=\Big\langle\sum_{k\in\mathbb N}\langle g_k,f_i\rangle g_k,f_j\Big\rangle=\langle f_i,f_j\rangle $$ as is readily verified again using the basis expansion in $\mathcal H$, so in total $G=V^\dagger V$. This concept of course still holds if one considers countably many vectors $(f_n)_{n\in\mathbb N}$ from $\mathcal H$, so the respective Gram matrix $G=(\langle f_i,f_j\rangle)_{i,j\in\mathbb N}$ is of infinite size - although the concept of the Gram determinant becomes a bit more involved in this case.

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_{Footnote: as a mathematical physicist, I'm used to the inner product being linear in the second argument. Of course, all of this still holds if the inner product is defined to be linear in the first argument as usual in pure mathematics, by changing $\langle g_i,x\rangle$ to $\langle x,g_i\rangle$ etc.}